logo

Logistics & Supply

A case study assignment in the field of Supply Chain Analysis and Design, where students are required to analyze and respond to a scenario, identify issues, and provide recommendations.

8 Pages1284 Words223 Views
   

Added on  2023-06-08

About This Document

This article discusses the optimal production schedule for a business in Logistics & Supply, including labor hours, hiring and firing costs, and total production costs. It also includes a chart showing the relationship between labor costs and the number of employees.

Logistics & Supply

A case study assignment in the field of Supply Chain Analysis and Design, where students are required to analyze and respond to a scenario, identify issues, and provide recommendations.

   Added on 2023-06-08

ShareRelated Documents
Running head: LOGISTICS & SUPPLY
Logistics & Supply
Name of the Student:
Name of the University:
Author’s Note:
Logistics & Supply_1
1LOGISTICS & SUPPLY
Table of Contents
Requirement 1:.................................................................................................................................2
Requirement 2:.................................................................................................................................4
Requirement 3:.................................................................................................................................6
Reference.........................................................................................................................................7
Logistics & Supply_2
2LOGISTICS & SUPPLY
Requirement 1:
It is assumed that the production for Month 1, Month 2, Month 2 and Month 4 are X1,
X2, X3 and X4 respectively and Z is total production cost for 4 months. On the basis of this
assumption, the Linear Programming Model to derive the optimal production level, which would
incur minimum cost, is developed below:
a) Total Nos. of Workers required:
Y1 = [4X1 – {20x(160+20)}]/(160+20) + 20
Y2 = [4X2 – {Y1x(160+20)}]/(160+20)
Y3 = [4X3 – {Y2x(160+20)}]/(160+20)
Y4 = [4X4 – {Y3x(160+20)}]/(160+20)
b) Closing Inventory:
S1 = X1 + 500 – 3000
S2 = X2 + S1 – 5000
S3 = X3 + S2 – 2000
S4 = X4 + S3 – 1000
c) Optimum Production Schedule:
Minimize Z=
[15X1 + 1500Y1 + (Y1x20x13) + 1600(Y1-20) + 2000(20-Y1) + 3S1] +
Logistics & Supply_3

End of preview

Want to access all the pages? Upload your documents or become a member.

Related Documents
BUS107e Qualitative Research Report
|9
|1309
|56